This topic is taught in Secondary 3 after introduction of Indices Law.
In solving indices equation involving the same base, one of the common techniques is by Substitution. But before you can do substitution, you need to apply indices law to ‘break down’ the equation. This process of breaking down is sometimes challenging for students. Knowing how to solve quadratic equation is also essential.
Sometimes, solving Indices Equation will also involve the concept of taking lg on both sides as well.
In the following example, you will Substitution and ‘Breaking down’ in action:
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I'm Ai Ling. I enjoy coaching students who have challenges with
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this is good
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You are such a great Mathematics. Iam impressed with your method of solving complex equations
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Yes you deserve a big up congratulation.Mathematics is no longer turf subject!
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how would you solve this if ^ means to the power of-
2^(x)3^(y)=6
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j hate Reply:
March 3rd, 2010 at 10:35 pm
with 2 unknowns you need 2 equations to solve for the unknowns
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how would you solve this if ^ means to the power of-
(2^2x+1)9^x=6^x
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wow this is really helpful, thank you soo much!
can someone help me with this? please
4^p-3 * 7^q-1 ?
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no root exist
Possible derivation:
d/dp(4^(p-3) 7^(q-1))
| Factor out constants:
= | 7^(q-1) (d/dp(4^(p-3)))
| Use the chain rule, d/dp(4^(p-3)) = ( d4^u)/( du) ( du)/( dp), where u = p-3 and ( d4^u)/( du) = 4^u log(4):
= | 7^(q-1) (4^(p-3) log(4) (d/dp(p-3)))
| Differentiate the sum term by term:
= | 4^(p-3) 7^(q-1) log(4) (d/dp(p)+d/dp(-3))
| The derivative of -3 is zero:
= | 4^(p-3) 7^(q-1) log(4) (d/dp(p)+0)
| The derivative of p is 1:
= | 1 4^(p-3) 7^(q-1) log(4)
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x5=125*squreroot of 2 pls solve d equation
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